Question

Find the general term and the sum of n terms of the series $1,0,1,8,29,80,193,....$

Answer 1

$\Rightarrow 10182980\mathrm{193....}$

$-1172151\mathrm{113....}$

$261430\mathrm{62....}$

$4816\mathrm{32....}$

$\therefore u_n=a.2^{n-1}+b{n}^2+cn+d$

Putting $n=1,2,3,4$ we get;-

$a=4,b=-1,c=-2,d=0$

$\therefore u_n=4.2^{n-1}-n^2-2n+0$

$=2^{n+1}-n^2-2n$

$S_n=\sum u_n$

$=(2^2+2^3+....+2^{n+1})-\sum n^2-2\sum n$

$=4(2^n-1)-\frac{1}{6}n(n+1)(2n+1)-n(n+1)$

$=2^{n+2}-4-\frac{1}{6}n(n+1)(2n+7)$